Question

y = sin^{-1} 2x \sqrt{1-x^{2}} dy/dx ज्ञात कीजिए

Answer 1

dy/dx =   2/√ (1 -  x²)  यदि y = Sin⁻¹(2x√(1 - x²))

Step-by-step explanation:

dy/dx ज्ञात कीजिए

y = Sin⁻¹(2x√(1 - x²))

=>Siny  = 2x√ (1 -  x²)

=> Cosy (dy/dx)  =   (2x(1/2)/√ (1 -  x²) )(-2x)  + 2√ (1 -  x²)

=> Cosy(dy/dx)  =  -2x²/√ (1 -  x²) + 2√ (1 -  x²)

=> Cosy(dy/dx)  =  (-2x²  +2 (1 - x²) )/ √ (1 -  x²)

=> Cosy(dy/dx)  =  2(1 - 2x²)/√ (1 -  x²)

Siny  =2x√ (1 -  x²)

Cosy = √ (1 - Sin²y)     = √( 1 - (2x√ (1 -  x²))²)

= √ (1  - 4x²(1 - x²))

=√ (4x⁴  - 4x² + 1)

= (1 - 2x²)

=> (1 - 2x²) dy/dx = 2(1 - 2x²)/√ (1 -  x²)

=> dy/dx =   2/√ (1 -  x²)

और अधिक जानें :

sin(x²+5)"

brainly.in/question/15286193

sin (ax+b) फलन का अवकलन कीजिए

brainly.in/question/15286166

Answer 2

Question :-

$\sf \: if \: y = sin^{-1}( 2x \sqrt{1-x^{2}}) \: \: then \: find \: \frac{dy}{dx}$

Answer :-

$\to \boxed{ \sf \frac{dy}{dx} = \frac{2}{ \sqrt{1 - {x}^{2} } } } \:$

Solution :-

We have ,

$\sf \longmapsto \: y = sin^{-1}( 2x \sqrt{1-x^{2}}) \: \\ \\ let \sf \: \: x = \sin \theta \: \\ \\ \: \to \: \theta \sf = { \sin}^{ - 1} x \\ \\ \bf hence \: \\ \\ \longmapsto \sf y = { \sin}^{ - 1} (2 \sin \theta \: \sqrt{1 - { \sin}^{2} \theta } ) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \because \: 1 - { \sin}^{2} \theta = { \cos}^{2} \theta} \\ \\ \to \sf y = { \sin}^{ - 1} \: (2 \sin \theta \: \sqrt{ { \cos}^{2} \theta} ) \\ \\ \to \sf \: y = { \sin}^{ - 1} (2 \sin \theta \cos \theta \: ) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \sf \sin2 \theta \: =2 \sin \theta \cos \theta} \\ \\ \to \sf \: y = { \sin}^{ - 1} ( \sin2 \theta) \\ \\ \sf \to \: y = 2 \theta \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \: \theta \sf = { \sin}^{ - 1} x} \\ \\ \to \sf y = 2 { \sin}^{ - 1}x \\ \\ \bf differentiate \: with \: respect \: to \: x \\ \\ \to \sf \frac{dy}{dx} = 2 \frac{d}{dx} { \sin}^{ - 1} x \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \sf \frac{d}{dx} { \sin}^{ - 1} x = \frac{1}{ \sqrt{1 - {x}^{2} } } } \\ \\ \to \boxed{ \sf \frac{dy}{dx} = \frac{2}{ \sqrt{1 - {x}^{2} } } }$